\(\int \frac {x^{-1+2 n}}{(a^2+2 a b x^n+b^2 x^{2 n})^{5/2}} \, dx\) [513]

Optimal result

Integrand size = 32, antiderivative size = 88 \[ \int \frac {x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2}} \, dx=\frac {a}{4 b^2 n \left (a+b x^n\right )^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {1}{3 b^2 n \left (a+b x^n\right )^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

1/4*a/b^2/n/(a+b*x^n)^3/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)-1/3/b^2/n/(a+b*x^n)^2/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1
/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {1369, 272, 45} \[ \int \frac {x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2}} \, dx=\frac {a}{4 b^2 n \left (a+b x^n\right )^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {1}{3 b^2 n \left (a+b x^n\right )^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[In]

Int[x^(-1 + 2*n)/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(5/2),x]

[Out]

a/(4*b^2*n*(a + b*x^n)^3*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) - 1/(3*b^2*n*(a + b*x^n)^2*Sqrt[a^2 + 2*a*b*x^n
+ b^2*x^(2*n)])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^4 \left (a b+b^2 x^n\right )\right ) \int \frac {x^{-1+2 n}}{\left (a b+b^2 x^n\right )^5} \, dx}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^n\right )\right ) \text {Subst}\left (\int \frac {x}{\left (a b+b^2 x\right )^5} \, dx,x,x^n\right )}{n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^n\right )\right ) \text {Subst}\left (\int \left (-\frac {a}{b^6 (a+b x)^5}+\frac {1}{b^6 (a+b x)^4}\right ) \, dx,x,x^n\right )}{n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ & = \frac {a}{4 b^2 n \left (a+b x^n\right )^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {1}{3 b^2 n \left (a+b x^n\right )^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45 \[ \int \frac {x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2}} \, dx=\frac {\left (-a-4 b x^n\right ) \left (a+b x^n\right )}{12 b^2 n \left (\left (a+b x^n\right )^2\right )^{5/2}} \]

[In]

Integrate[x^(-1 + 2*n)/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(5/2),x]

[Out]

((-a - 4*b*x^n)*(a + b*x^n))/(12*b^2*n*((a + b*x^n)^2)^(5/2))

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.42

method	result	size
risch	\(-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, \left (4 b \,x^{n}+a \right )}{12 \left (a +b \,x^{n}\right )^{5} b^{2} n}\)	\(37\)

[In]

int(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/12*((a+b*x^n)^2)^(1/2)/(a+b*x^n)^5*(4*b*x^n+a)/b^2/n

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2}} \, dx=-\frac {4 \, b x^{n} + a}{12 \, {\left (b^{6} n x^{4 \, n} + 4 \, a b^{5} n x^{3 \, n} + 6 \, a^{2} b^{4} n x^{2 \, n} + 4 \, a^{3} b^{3} n x^{n} + a^{4} b^{2} n\right )}} \]

[In]

integrate(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(5/2),x, algorithm="fricas")

[Out]

-1/12*(4*b*x^n + a)/(b^6*n*x^(4*n) + 4*a*b^5*n*x^(3*n) + 6*a^2*b^4*n*x^(2*n) + 4*a^3*b^3*n*x^n + a^4*b^2*n)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(x**(-1+2*n)/(a**2+2*a*b*x**n+b**2*x**(2*n))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2}} \, dx=-\frac {4 \, b x^{n} + a}{12 \, {\left (b^{6} n x^{4 \, n} + 4 \, a b^{5} n x^{3 \, n} + 6 \, a^{2} b^{4} n x^{2 \, n} + 4 \, a^{3} b^{3} n x^{n} + a^{4} b^{2} n\right )}} \]

[In]

integrate(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(5/2),x, algorithm="maxima")

[Out]

-1/12*(4*b*x^n + a)/(b^6*n*x^(4*n) + 4*a*b^5*n*x^(3*n) + 6*a^2*b^4*n*x^(2*n) + 4*a^3*b^3*n*x^n + a^4*b^2*n)

Giac [F]

\[ \int \frac {x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2}} \, dx=\int { \frac {x^{2 \, n - 1}}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(5/2),x, algorithm="giac")

[Out]

integrate(x^(2*n - 1)/(b^2*x^(2*n) + 2*a*b*x^n + a^2)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2}} \, dx=\int \frac {x^{2\,n-1}}{{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{5/2}} \,d x \]

[In]

int(x^(2*n - 1)/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(5/2),x)

[Out]

int(x^(2*n - 1)/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(5/2), x)